Question

Consider a large cubic ice block floating in water with top surface of the ice block parallel to the surface of the sea. The specific gravities of ice and seawater are 0.9 and 1.0, respectively. If a 20-cm-high portion of the ice block extends above the surface of the water, determine the height of the ice block below the surface.

• A. 0.9 m
• B. 1.0 m
• C. 1.6 m
• D. 1.8 m

Hint:

For any floating body of uniform cross-section, the fraction of volume submerged is equal to the ratio of specific gravities: $V_{\text{sub}}/V_{\text{total}} = S_{\text{body}}/S_{\text{fluid}}$.
Here, $90\%$ of the block is submerged and $10\%$ is above water.
If $10\%$ corresponds to $0.2\text{ m}$, then the total height is $2.0\text{ m}$, meaning $1.8\text{ m}$ is submerged.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks to find the height of a floating cubic ice block that remains submerged below the water level.
We are given the specific gravities of both the floating body and the fluid, as well as the height of the block projecting above the water line.

Step 2: Key Formula or Approach:
According to Archimedes' Principle, the weight of the floating body equals the weight of the fluid displaced by its submerged volume:
\[ W_{\text{body}} = F_{\text{buoyancy}} \] \[ \rho_{\text{ice}} \cdot g \cdot V_{\text{total}} = \rho_{\text{water}} \cdot g \cdot V_{\text{submerged}} \] For a block of uniform cross-sectional area \(A\) and total height \(H\):
\[ S_{\text{ice}} \cdot H = S_{\text{water}} \cdot h_{\text{sub}} \] where \(h_{\text{sub}}\) is the submerged height, and \(H = h_{\text{sub}} + h_{\text{above}}\).

Step 3: Detailed Explanation:

• Identify the given parameters:
Specific gravity of ice, \(S_{\text{ice}} = 0.9\).
Specific gravity of seawater, \(S_{\text{water}} = 1.0\).
Height above water, \(h_{\text{above}} = 20\text{ cm} = 0.2\text{ m}\).

• Express the total height of the ice block as:
\[ H = h_{\text{sub}} + 0.2 \]
• Substitute this expression into the equilibrium equation:
\[ 0.9 \times (h_{\text{sub}} + 0.2) = 1.0 \times h_{\text{sub}} \] \[ 0.9 h_{\text{sub}} + 0.18 = h_{\text{sub}} \]
• Solve for the submerged height \(h_{\text{sub}}\):
\[ h_{\text{sub}} - 0.9 h_{\text{sub}} = 0.18 \] \[ 0.1 h_{\text{sub}} = 0.18 \implies h_{\text{sub}} = \frac{0.18}{0.1} = 1.8\text{ m} \]

Step 4: Final Answer:
The height of the ice block below the surface is \(1.8\text{ m}\).